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Functions Test 20

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Functions Test 20
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  • Question 1
    1 / -0
    The domain of the function, $$f(x) = \displaystyle \dfrac {1}{\sqrt{[x]^2-[x]-6}}$$  is
    Solution
    Given : $$f(x) = \displaystyle \frac {1}{\sqrt{[x]^2-[x]-6}}$$

    $$f(x)$$ is real if $${\sqrt{[x]^2-[x]-6}}\in R$$ 

    i.e.
     $$[x]^{ 2 }-[x]-6>0$$
    $$\implies [x]^{ 2 }-3[x]+2[x]-6>0$$
    $$\implies [x]([x]-3)+2([x]-3)>0$$
    $$\implies ([x]-3)([x]+2)>0$$
    $$ \implies \left[ x \right] <-2 $$ and $$\left[ x \right] >3$$

    Therefore, $$x \in (-\infty , -3] \cup [4, \infty )$$
  • Question 2
    1 / -0
    The domain of the function $$\ln (x-1)$$ is.
    Solution
    $$\log(x-1)$$
    where $$ (x-1)>0$$
    $$x>1$$
    Domain of $$\log(x-1)$$  $$=(1,\infty)$$
  • Question 3
    1 / -0
    The domain of the function, $$f(x) = \log_{10} (\sqrt{x-4}+\sqrt {6-x})$$ is
    Solution
    Given,
    $$f(x) = \log_{10} (\sqrt{x-4}+\sqrt {6-x})$$

    For given function to be defined,
    $$\Rightarrow x-4 \geq  0 \,$$ and $$\, 6-x \geq  0 \Rightarrow  x \geq  4 \, $$and $$ \,x \leq  6$$

    $$\therefore $$  Domain of $$(x) = [4, 6]$$
  • Question 4
    1 / -0
    If $$f : [0, \Pi ] \rightarrow  [-1, 1]$$, f(x) = cosx, then f is.
    Solution
    f takes all values from $$[-1,1]$$ while travelling from $$[0,\pi]$$. None of the values are repeated either. 

    Hence it is one-one and onto.
  • Question 5
    1 / -0
    If $$f(x) = \left\{\begin{matrix} 1&x \in Q \\ 0 &x \notin  Q\end{matrix}\right.$$ then $$fof(\sqrt 3 )$$ is equal to
    Solution
    We know $$\sqrt{3}$$ is an irrational number

    Thus $$f(\sqrt{3}) =0$$ and $$0$$ is a rational number

    Hence $$fof(\sqrt 3 )= f(0) = 1$$
  • Question 6
    1 / -0
    The domain of the function $$\log \displaystyle \sqrt {\frac {3-x}{2}}$$ is.
    Solution
    We have $$f(x) = \log \sqrt {\dfrac {3-x}{2}}$$

    $$\because     \dfrac {3-x}{2}>0$$

    $$\Rightarrow 3 -x > 0$$

    $$\Rightarrow x < 3 $$

    $$\Rightarrow x \in (-\infty , 3)$$

    $$\therefore $$   domain of $$f(x)$$ is $$(-\infty , 3)$$.
  • Question 7
    1 / -0
    The domain of the function $$f(x) = \displaystyle \cfrac { 1 }{ \sqrt { x-\left[ x \right]  }  } $$
    Solution
    We have $$f(x) =\dfrac {1} {\sqrt{x-|x|}}$$

    $$\Rightarrow   x -[x] > 0$$

    $$\Rightarrow   x > [x]$$ is defined only when $$x\notin   Z.$$

    Domain of $$f(x)$$ is $$R-Z$$
  • Question 8
    1 / -0
    The domain of the function $$\displaystyle f(x) =\frac {1}{\sqrt {(x-1) (x-2)}}$$ is.
    Solution
    For $$f(x)$$ to be defined $$(x-1)(x-2) > 0$$
    $$\Rightarrow x \in (-\infty, 1) \cup (2, \infty)$$
  • Question 9
    1 / -0
    The domain of $$f(x) = \log_e |\log_ex|$$ is.
    Solution
    $$f(x) = log|log x|, (x)$$ is defined if $$|log x| > 0$$ and $$x > 0$$ 

    $$i.e., if x > 0$$ and $$x$$ $$\neq$$ 1

    $$(\because |log x| > 0\,\, if\, \,  x \neq  1)$$

    $$\Rightarrow  x \in (0, 1) \cup  (1, \infty )$$.
  • Question 10
    1 / -0
    If functions $$f\left ( x \right )$$ and $$g\left ( x \right )$$ are defined on $$R\rightarrow R$$ such that
    $$f(x)=x+3, x$$ $$\in  $$ rational
             $$ =4x, x$$ $$\in $$ irrational
    $$g(x)=x+\sqrt{5}$$, x$$\in $$ irrational
          $$  =-x, x$$ $$\in $$ rational
    then $$\left ( f-g \right )\left ( x \right )$$ is
    Solution
    $$f\left( x \right) =x+3, x\epsilon$$  rational 
              $$=4x, x\epsilon$$  irrational

    $$g\left( x \right) =x+\sqrt { 5 } , x \epsilon$$  irrational
              $$= -x,x\epsilon$$  rational 

    $$(f-g)(x)=2x+3, x\ \epsilon$$  rational
                         $$+3x-\sqrt { 5\quad  } x\quad \epsilon$$ irrational

    For one-one; 
    We know that one one function is a $${ f }^{ \underline { n }  }$$ for which every element of the range of the function corresponds to exactly one element of the domain.
     
    But in this case this is not true as for all rational nos. the$$ { f }^{ \underline { n }  }$$ is one one but for irrational $${ f}^{ \underline { n }  }$$, every elements of the range does not corresponds to exactly one element of the domain. 

    $$(f-g)(x)\neq  (f-g)({ x }^{ \prime  })$$         when $$x \epsilon$$ rational and 
                                                                   $$x \epsilon$$ irrational
    for onto:
    The $${ f }^{ \underline { n }  }(f-g)(x)$$ does not cover the whole range of the function.
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